Equação de Estado van der Waals

Prévia do material em texto

SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 25
With the data given
b = Vm −
RTV 2m
pV 2m + a
= (4.00 × 10−4 m3mol−1)
− (8.3145 JK−1mol−1) × (288 K) × (4.00 × 10−4 m3mol−1)2
(4.0 × 106 Pa) × (4.00 × 10−4 m3mol−1)2 + (0.76 m6 Pamol−2)
= 1.3 × 10−4 m3mol−1
where 1 Pa = 1 kgm−1 s−2 and 1 J = 1 kgm2 s−2 have been used.
�e compression factor Z is de�ned in [1C.1–20] as Z = Vm/V○m, where V○m is
the molar volume of a perfect gas under the same conditions. �is volume is
computed from the equation of state for a perfect gas, [1A.4–8], as V○m = RT/p,
hence Z = pVm/RT , [1C.2–20]. With the data given
Z = pVm
RT
= (4.0 × 106 Pa) × (4.00 × 10−4 m3mol−1)
(8.3145 JK−1mol−1) × (288 K)
= 0.67
Solutions to problems
P1C.2 (a) Using the perfect gas law, pV = nRT , the molar volume is calculate as
Vm = RT
p
= (8.2057 × 10−2 dm3 atmK−1mol−1) × (350 K)
2.30 atm
= 12.5 dm3mol−1
(b) �e van derWaals equation of state in terms of the molar volume is given
by [1C.5b–24], p = RT/(Vm−b)−a/V 2m; the term a/V 2m is due to attractive
interactions.
�is equation is a cubic inVm which can be solved numerically. A simpler
approach is to approximate a/V 2m as a/(V○m)2, where V○m is the molar
volume of a perfect gas under the prevailing conditions. �e van der
Waals equation is then rearranged to give a simpler expression for Vm
p = RT
(Vm − b)
− a
(V○m)2
hence Vm = RT
p + [a/(V○m)2]
+ b
�e van der Waals constants for Cl2 are a = 6.260 atmdm6mol−2 and
b = 5.42 × 10−2 dm3mol−1. With these values, the �rst approximation to
the molar volume is calculated as
Vm = (8.2057 × 10−2 dm3 atmK−1mol−1) × (350 K)
(2.30 atm) + [(6.260 atmdm6mol−2)/(12.5 dm3mol−1)2]
+ 5.42 × 10−2 dm3mol−1 = 12.3 dm3mol−1
�is approximate value forVm is then used in place ofV○m and the process
repeated, to give Vm = 12.3 dm3mol−1 – the same value.�e process has
converged (to this level of precision).